OXIDATIONREDUCTION

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• Chapter 22
• OXIDATION-REDUCTION
• REACTIONS

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• Oxidation as you might know it involves the
  burning of oxygen during combustion reactions.
• Not all oxidation processes involve burning.
• For example, bleaching is an example of oxidation
  that does not involve burning.
• Neither does rusting involve burning, but it is
  an oxidation process.
• The opposite of oxidation is reduction which
  originally was considered the removal of oxygen.
• Oxidation and reduction always occur
  simultaneously.

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• ELECTRON TRANSFER IN REDOX REACTIONS
• Today the concepts of oxidation-reduction
  reactions have been extended to those that dont
  even involve oxygen.
• Oxidation- complete or partial loss of electrons
  or the gain of oxygen.
• Reduction- complete or partial gain of electrons
  or loss of oxygen
• For example
• Mg S ? Mg2 S2-
• Transfer of 2 electrons to sulfur.
• Sulfur is reduced and magnesium is oxidized

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• The substance that loses the electrons is called
  the reducing agent.
• The substance that gains the electrons is called
  the oxidizing agent.
• What is oxidized and what is reduced in this
  single replacement reaction?
• 2AgNO3(aq) Cu(s)? Cu(NO3)2(aq) 2 Ag(s)
• Step 1
• Rewrite the equation showing the ions.
• (2Ag 2NO3- ) Cu ? ( Cu2 2NO3-) 2Ag
• Oxidation Cu ? Cu2 2e- reducing agent
• Reduction 2Ag 2e- ? 2Ag oxidizing agent

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• Easy to tell ionic situations and transfer of
  electrons.
• What about covalent compounds?
• In a covalent situation there is not a complete
  transfer of electrons.
• Electron shells are shared rather than an
  electron transfer.
• Water, for example, is a covalent situation but
  the electrons are pulled toward the oxygen atom.
• The result is a shift of electrons
• Hydrogen is oxidized because it undergoes a
  partial loss of electrons and oxygen is reduced
  because of the shift of electrons toward it.

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• CORROSION
• In industry, billions of dollars a year are spent
  in the prevention/repair of corrosion.
• The process of corrosion is increased in humid
  conditions, and is greatly increased in the
  presence of salts and acids.
• Why do you think that acids and salts would
  enhance the corrosive process?
• If their ions are freed up by the process of
  dissolving(humidity allows dissolving to occur)
  then the conduction of electron transfer takes
  place with more ease.

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• Not all metals corrode easily.
• Gold and platinum are called noble metals
  because they are very resistant to losing their
  electrons.
• If an oxide coating occurs on the surface of the
  metals atoms, then it protects it from further
  corrosion underneath the atom.
• For example, aluminum oxidizes so readily that
  very tightly packed aluminum oxide particles form
  and protect it from further corrosion.
• Coatings are used to protect metals from
  corrosion by preventing water molecules in.
• Another method used to prevent rusting is
  introduce another metal to the mix that actually
  oxidizes more readily. Magnesium, for example
  oxidizes more readily than iron.
• As a result the magnesium serves as an electric
  current where it offers the electrons back to the
  iron and reduces the iron to a neutral atom.

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• Open up your text to page 653 and work on
  problems 3 to 8

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• 3. Define oxidation and reduction in terms of
• a. gain or loss of electrons oxidation would be
  a loss of electrons and reduction would be a gain
  of electrons.
• b. gain or loss of hydrogen(covalent bond)-
  oxidation would be a loss of hydrogen and
  reduction would be a gain of hydrogen.
• c. gain or loss of oxygen- oxidation would be the
  gain of oxygen and reduction would be the loss of
  oxygen.
• d. shift of electrons(covalent bond)- oxidation
  would be a shift of electrons away from atom and
  reduction would be a shift of electron toward the
  atom

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• 4. State the characteristics of a redox reaction,
  and explain how to identify the oxidizing agent
  and the reducing agent.
• There is a partial or complete transfer of
  electrons between a reducing agent, which is
  oxidized, and oxidizing agent, which is reduced.
• 5. Which of the following would most likely be
  oxidizing agents and which would most likely be
  reducing agents?
• a. Cl2 - oxidizing agent
• b. K -reducing agent
• c. Ag- oxidizing agent

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• 6. Determine which reactant is oxidized and which
  is reduced and which is the reducing agent and
  which is the oxidizing agent. Refer to table 14.2
  on page 405.
• a. H2(g) Cl2 (g) ? 2HCL(g)
• H2 oxidized Cl2 reduced
• b. S(s) Cl2 (g)? SCl2(g)
• S oxidized Cl2 reduced
• c. N2 2O2?2NO2
• N2 oxidized O2 reduced
• d. 2Li F2 ? 2LiF
• Li oxidized F2 reduced
• e. H2 S ? H2S
• H2 oxidized S reduced

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• 7. Use electron transfer or shift to identify
  what is oxidized and what is reduced in each
  reaction. Make use of electronegativity values,
  as needed, for molecular compounds.
• a. 2Na(s) Br2(l) ? 2NaBr(s)
• Na oxidized Br2 reduced
• b. N2 (g) 3H2 (g) ? 2NH3 (g)
• H2 oxidized N2 reduced
• c. S(s) O2(g) ? SO2(g)
• S oxidized O2 reduced( S lt electronegative than
  O)
• d. Mg(s) Cu(NO3)2(aq)?Mg(NO3)2(aq) Cu(s)
• Mg Oxidized Cu2 reduced

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• 8. Why would a metal corrode more quickly in salt
  water than in distilled water?
• Ions make electron transfer easier.

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• ASSIGNING OXIDATION NUMBERS
• An oxidation number is a positive or negative
  number assigned to an atom that depends on a set
  of rules for that particular atom.
• Unlike the charge of an ion, the oxidation number
  is put after the charge. The charge of an ion is
  put after the number.
• NaCl, for example, the charge of each ion is 1.
• In this particular case, the oxidation number is
  the same number as the charge.
• So the oxidation number for sodium is 1 and for
  chlorine is a -1.

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• Rules for assigning oxidation numbers
• 1. for monatomic ions, the oxidation charge and
  number is the same as the ion.
• 2. oxidation number for hydrogen in a compound is
  1 except in metal oxides like NaH, where it is
  then a -1.
• 3. oxidation number for oxygen in compounds is a
  -2 except for cases like H2O2 where it is a -1.
• 4. If a metal is not combining with anything
  else, it is said to be in elemental form. A
  diatomic molecule is also considered elemental
  form. The oxidation number in these circumstances
  is 0.

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• 5. For any neutral compound, the sum of their
  oxidation numbers must equal 0.
• 6. For polyatomic ions, the oxidation number is
  also equal to the overall charge. The oxidation
  numbers of each element in the molecule depends
  on electronegativity influences. Flourine being
  the most electronegative.
• What is the oxidation numbers for each kind of
  atom in the following compound?
• SO2
• Remember that Oxygen is more electronegative.
• Sulfur will be a 4 and oxygen will be a -2.

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• OXIDATION NUMBER CHANGES IN CHEMICAL REACTIONS
• A decrease in oxidation number during a chemical
  reaction will indicate that the atom was reduced.
• An increase in the oxidation number will indicate
  that an atom was oxidized.
• For example
• 1 5 -2 0
  2 5 -2 0
• 2AgNO3(aq) Cu(s) ? Cu(NO3)2(aq) 2AG(s)
• In this reaction the oxidation number of silver
  decreases from 1 to 0, which indicates
  reduction.
• Copper is oxidized because it went from 0 to 2.

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• Where did the oxidation number of 5 for nitrogen
  come from?
• Nitrogen has 5 valence electrons and has room for
  3 more electrons. When it forms a molecule with
  oxygen it bonds to 3 oxygen atoms by sharing
  electron shells with each oxygen atom.
• Oxygen has 6 valence electrons. Room for 2 more .
• There are a total of 3 oxygen in all, each
  carrying a -2 oxidation number.
• Because oxygen is more electronegative than
  nitrogen, oxygen wins the tug of war for the
  charge in the covalent situation. Collectively,
  the oxygen gains the charge advantage for each
  atom. Totaling -6
• As a result more of the negative charge is
  favored toward the oxygen atoms in the molecule
  and nitrogen loses the electrons somewhat. The
  nitrogen atom is oxidized and oxygen is reduced.

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• In the case where there is no charge, prior to
  NO3 responding to the metal to form the ionic
  bond, the oxidation number for nitrogen is a 6.
• Now that it is responding to the metal(Ag) it is
  taking an electron from Ag.
• Remember when an ionic bond is forming, a
  transfer of electrons occurs.
• Ag has a 1 charge and the polyatomic ion has the
  overall charge of a 1-.
• That leaves nitrogen with a different oxidation
  number.
• It no longer is 6, it is given a 5 charge
  instead.
• Typically oxygens oxidation number remains the
  same and the gain of the charge changes the other
  atom bound to oxygen.

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• Use the changes in oxidation numbers to identify
  which atoms are oxidized and which atoms are
  reduced in the following reaction
• Cl2(g) 2 HBr(aq) ?2HCl(aq) Br2(l)
• 0 1 -1 1-1
  0
• Cl2(g) 2 HBr(aq) ?2HCl(aq) Br2(l)
• Chlorine is reduced, bromide ion is oxidized.
• Try this at your desk
• C(s) O2(g) ? CO2(g)
• 0 0 4 -2
• C(s) O2(g) ? CO2(g)
• Carbon is oxidized and oxygen is reduced

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• When assigning oxidation numbers, refer to the
  rules
• Remember that hydrogen is always a 1 (except for
  when it is a metal hydride (binary situation)
  where it will be more electronegative than the
  metal present.
• The following example demonstrates H with a 1
  oxidation number.
• Zn(s) 2MnO2(s) 2NH4Cl(aq) ?
• ZnCl2(aq) Mn2O3(s) 2NH3(g)
  H2O(l)
• 0 4 -2 -3 1
  -1
• Zn(s) 2MnO2(s) 2NH4Cl(aq) ?
• 2-1 3
  -2 -3 1 1 -2
• ZnCl2(aq) Mn2O3(s) 2NH3(g)
  H2O(l)

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• 0 4 -2 -3 1
  -1
• Zn(s) 2MnO2(s) 2NH4Cl(aq) ?
• 2-1 3
  -2 -3 1 1 -2
• ZnCl2(aq) Mn2O3(s) 2NH3(g)
  H2O(l)
• How did I know how to choose the correct number?
• Zn, elemental form, so it is zero
• O -2 and since there are 2 Oxygens, the Mn must
  be a 4.
• For NH4, You know H 1 so before it is charged,
  N has a -4.
• As an ion, N loses an electron and then becomes
  a -3.
• Remember the overall charge must equal its
  polyatomic charge of 1
• Zinc is oxidized(0 to 2) and manganese is
  reduced(4 to 3)

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• Do questions
• 13 to 16 at your desk on page 659

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• 13. (a) What is the oxidation number of nitrogen
  in nitrogen gas? Explain
• The oxidation number of an uncombined atom is
  zero.
• (b)How would you determine the oxidation numbers
  of an element in a compound?
• The sum of the oxidation numbers must equal zero.
  If you know one then the other can be calculated.

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• (c) How is charge used to determine the oxidation
  numbers to the elements in a polyatomic ion?
• The sum of the oxidation numbers must equal the
  charge on the polyatomic ion.
• 14. How are oxidation numbers determined and
  used?
• Oxidation s are assigned by rules and
  electronegativity values and are used to
  determine oxidizing and reducing agents.

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• 15. Use the charges in oxidation numbers to
  identify which atoms are oxidized and which atoms
  are reduced in each reaction.
• 0 0 1 -1
• a. 2Na(s) Cl2(g) ? 2NaCl(s)
• Na oxidized Cl2 reduced
• 1 5-2 1-1 2-2
  0
• b. 2HNO3(aq) 6HI(aq) ? 2NO(g) 3I2(s) 4 H2O
• I is oxidized and N is reduced
• 1-2 1 5-2 0
  2-2 1 -2
• c. 3H2S(aq) 2HNO3(aq) ? 3S(s) 2NO(g) 4H2O(l)
• S is oxidized and N is reduced.

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• d.
• 2 6 -2 1 -2 0
  4 -2 16 -2
• 2PbSO4(s) 2H2O(l) ?Pb(s) PbO2(s) 2H2SO4(aq)
• Pb in PbSO4 is both the oxidizing and reducing
  agent.
• This is determined by examining the oxidation
  numbers.
• Pb went from 2 to 0 reduced (gained electrons)
• Pb went from 2 to 4 oxidized (lost electrons)
• Sulfur, oxygen, and hydrogen remained the same on
  both sides of the equation.

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• 16. Identify the oxidizing agent and the reducing
  agent in each of the reactions from 15.
• a. Na reducing agent and Cl2 oxidizing agent
• b. N in HNO3 oxidizing agent I in HI is the
  reducing agent.
• c. N in HNO3 oxidizing agent and S in H2S is the
  reducing agent.

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• IDENTIFYING REDOX REACTIONS
• How can you tell if a redox reaction has
  occurred?
• If you apply oxidation numbers to the reactants
  and products of a chemical reaction and there is
  no change in any of the components oxidation
  numbers, then a redox did not occur.
• For example double-replacement and acid-base
  reactions are not redox reactions.
• Nitrogen(II)oxide is formed during an electrical
  storm in the atmosphere when N2 and O2 molecules
  combine.
• N2(g) O2(g) ? 2NO(g)
• Is this a redox reaction?
• Assign oxidation numbers and find out.

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• Yes Nitrogen went from 0 to 2 (oxidized)
• Oxygen went from 0 to -2 (reduced)
• Various other changes signal an
  oxidation-reduction reaction.
• One such change is a color change.
• In the following single-replacement reaction you
  can see a dramatic color change when the copper
  replaces silver in a solution of AgNO3.
• Does the solutions color look familiar to you?

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• Coefficients are not used in determining
  oxidation numbers however, they are considered
  when balancing a redox reaction.
• Rules for balancing a redox reaction
• 1. Assign appropriate oxidation numbers to a
  skeleton equation.
• 3 -2 2-2 0
  4-2
• Example Fe2O3 CO ? Fe CO2
• 2. Identify which atoms are oxidized and which
  atoms are reduced. Carbon-oxidized and
  Iron-reduced.

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• 
  2
• 3 -2 2-2 0
  4-2
• Fe2O3 CO ? Fe CO2
• -3
• 3. Use one bracketing line to connect the atoms
  that undergo oxidation, and another to those that
  undergo reduction. Write the oxidation number
  change on the midpoint of each line.

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• 4.Make the total increase in oxidation equal to
  the total decrease in oxidation by using
  appropriate coefficients.
• 
  3 X (2) 6

Fe2O3 3CO ? 2Fe 3CO2
2 X (-3) -6
By multiplying the net oxidation by 3 and net
reduction by 2, the equation can be balanced
with the correct coefficients. A 3 is placed in
front of CO and CO2. A 2 in front of Fe, not
necessary to place a 2 in front of Fe2O3 since Fe
already has 2 atoms.
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• Finally
• Step 5
• Make sure that the equation is balanced for both
  the number of atoms and for charge.

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• WORK ON PROBLEM 26 ON PAGE 673
• Problem 26
• a,b,and c
• using the oxidation-number-change method.

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• 26. Balance each redox equation, using the
  oxidation-number-change method.
• a. Ba (s) O2(g) ? BaO (s)
• 0 0 2 -2
• . Ba (s) O2(g) ? BaO (s)
• . Ba (s) O2(g) ? BaO (s)
• -2
• Ba (s) O2(g) ?BaO (s)
• 2

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• In this case the magnitude of both changes are
  equal, but the equation is not . By multiplying
  both
• numbers by two and changing the coefficients to
  2,
• Balances the equation and oxidation numbers .
• 2 Ba (s) O2(g) ? 2 BaO (s)

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• HOMEWORK PACKETS FOR CHAPTER 22 DUE WEDNESDAY
• APRIL 8, 20
• TEST ON TUESDAY
• APRIL 14, 2008